Given the function f (x) = xlnx, if f (x) > = AX-1 holds for any x > 0, then the value range of a is a = 1

Given the function f (x) = xlnx, if f (x) > = AX-1 holds for any x > 0, then the value range of a is a = 1

B
a=(xlnx+1)/x
=lnx+1/x
The derivative is (x-1) / x ^ 2
When x = 1, there is a minimum value of 1
So choose B

Given the function f (x) = xlnx, G (x) = 2x-3. (1) prove that f (x) > G (x)

Let H (x) = f (x) - G (x) = xlnx-2x + 3 (domain x > 0)
Derivation H '(x) = LNX + 1-2 = lnx-1
Let H '(x) = 0 get x = e, and the second derivative H' '(x) = 1 / x > 0
That is, H (E) is the minimum, H (x) > = H (E) = e-2e + 3 = 3-E > 0 (x > 0)
So f (x) > G (x)